-0.000 000 000 742 147 672 54 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 672 54(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 742 147 672 54(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 742 147 672 54| = 0.000 000 000 742 147 672 54


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 147 672 54.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 742 147 672 54 × 2 = 0 + 0.000 000 001 484 295 345 08;
  • 2) 0.000 000 001 484 295 345 08 × 2 = 0 + 0.000 000 002 968 590 690 16;
  • 3) 0.000 000 002 968 590 690 16 × 2 = 0 + 0.000 000 005 937 181 380 32;
  • 4) 0.000 000 005 937 181 380 32 × 2 = 0 + 0.000 000 011 874 362 760 64;
  • 5) 0.000 000 011 874 362 760 64 × 2 = 0 + 0.000 000 023 748 725 521 28;
  • 6) 0.000 000 023 748 725 521 28 × 2 = 0 + 0.000 000 047 497 451 042 56;
  • 7) 0.000 000 047 497 451 042 56 × 2 = 0 + 0.000 000 094 994 902 085 12;
  • 8) 0.000 000 094 994 902 085 12 × 2 = 0 + 0.000 000 189 989 804 170 24;
  • 9) 0.000 000 189 989 804 170 24 × 2 = 0 + 0.000 000 379 979 608 340 48;
  • 10) 0.000 000 379 979 608 340 48 × 2 = 0 + 0.000 000 759 959 216 680 96;
  • 11) 0.000 000 759 959 216 680 96 × 2 = 0 + 0.000 001 519 918 433 361 92;
  • 12) 0.000 001 519 918 433 361 92 × 2 = 0 + 0.000 003 039 836 866 723 84;
  • 13) 0.000 003 039 836 866 723 84 × 2 = 0 + 0.000 006 079 673 733 447 68;
  • 14) 0.000 006 079 673 733 447 68 × 2 = 0 + 0.000 012 159 347 466 895 36;
  • 15) 0.000 012 159 347 466 895 36 × 2 = 0 + 0.000 024 318 694 933 790 72;
  • 16) 0.000 024 318 694 933 790 72 × 2 = 0 + 0.000 048 637 389 867 581 44;
  • 17) 0.000 048 637 389 867 581 44 × 2 = 0 + 0.000 097 274 779 735 162 88;
  • 18) 0.000 097 274 779 735 162 88 × 2 = 0 + 0.000 194 549 559 470 325 76;
  • 19) 0.000 194 549 559 470 325 76 × 2 = 0 + 0.000 389 099 118 940 651 52;
  • 20) 0.000 389 099 118 940 651 52 × 2 = 0 + 0.000 778 198 237 881 303 04;
  • 21) 0.000 778 198 237 881 303 04 × 2 = 0 + 0.001 556 396 475 762 606 08;
  • 22) 0.001 556 396 475 762 606 08 × 2 = 0 + 0.003 112 792 951 525 212 16;
  • 23) 0.003 112 792 951 525 212 16 × 2 = 0 + 0.006 225 585 903 050 424 32;
  • 24) 0.006 225 585 903 050 424 32 × 2 = 0 + 0.012 451 171 806 100 848 64;
  • 25) 0.012 451 171 806 100 848 64 × 2 = 0 + 0.024 902 343 612 201 697 28;
  • 26) 0.024 902 343 612 201 697 28 × 2 = 0 + 0.049 804 687 224 403 394 56;
  • 27) 0.049 804 687 224 403 394 56 × 2 = 0 + 0.099 609 374 448 806 789 12;
  • 28) 0.099 609 374 448 806 789 12 × 2 = 0 + 0.199 218 748 897 613 578 24;
  • 29) 0.199 218 748 897 613 578 24 × 2 = 0 + 0.398 437 497 795 227 156 48;
  • 30) 0.398 437 497 795 227 156 48 × 2 = 0 + 0.796 874 995 590 454 312 96;
  • 31) 0.796 874 995 590 454 312 96 × 2 = 1 + 0.593 749 991 180 908 625 92;
  • 32) 0.593 749 991 180 908 625 92 × 2 = 1 + 0.187 499 982 361 817 251 84;
  • 33) 0.187 499 982 361 817 251 84 × 2 = 0 + 0.374 999 964 723 634 503 68;
  • 34) 0.374 999 964 723 634 503 68 × 2 = 0 + 0.749 999 929 447 269 007 36;
  • 35) 0.749 999 929 447 269 007 36 × 2 = 1 + 0.499 999 858 894 538 014 72;
  • 36) 0.499 999 858 894 538 014 72 × 2 = 0 + 0.999 999 717 789 076 029 44;
  • 37) 0.999 999 717 789 076 029 44 × 2 = 1 + 0.999 999 435 578 152 058 88;
  • 38) 0.999 999 435 578 152 058 88 × 2 = 1 + 0.999 998 871 156 304 117 76;
  • 39) 0.999 998 871 156 304 117 76 × 2 = 1 + 0.999 997 742 312 608 235 52;
  • 40) 0.999 997 742 312 608 235 52 × 2 = 1 + 0.999 995 484 625 216 471 04;
  • 41) 0.999 995 484 625 216 471 04 × 2 = 1 + 0.999 990 969 250 432 942 08;
  • 42) 0.999 990 969 250 432 942 08 × 2 = 1 + 0.999 981 938 500 865 884 16;
  • 43) 0.999 981 938 500 865 884 16 × 2 = 1 + 0.999 963 877 001 731 768 32;
  • 44) 0.999 963 877 001 731 768 32 × 2 = 1 + 0.999 927 754 003 463 536 64;
  • 45) 0.999 927 754 003 463 536 64 × 2 = 1 + 0.999 855 508 006 927 073 28;
  • 46) 0.999 855 508 006 927 073 28 × 2 = 1 + 0.999 711 016 013 854 146 56;
  • 47) 0.999 711 016 013 854 146 56 × 2 = 1 + 0.999 422 032 027 708 293 12;
  • 48) 0.999 422 032 027 708 293 12 × 2 = 1 + 0.998 844 064 055 416 586 24;
  • 49) 0.998 844 064 055 416 586 24 × 2 = 1 + 0.997 688 128 110 833 172 48;
  • 50) 0.997 688 128 110 833 172 48 × 2 = 1 + 0.995 376 256 221 666 344 96;
  • 51) 0.995 376 256 221 666 344 96 × 2 = 1 + 0.990 752 512 443 332 689 92;
  • 52) 0.990 752 512 443 332 689 92 × 2 = 1 + 0.981 505 024 886 665 379 84;
  • 53) 0.981 505 024 886 665 379 84 × 2 = 1 + 0.963 010 049 773 330 759 68;
  • 54) 0.963 010 049 773 330 759 68 × 2 = 1 + 0.926 020 099 546 661 519 36;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 742 147 672 54(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 672 54(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 672 54(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 672 54 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111